The pH of 0.01 M NaOH (aq) solution will be

(1) 9

(2) 7.01

(3) 2

(4) 12

Correct Answer: (4) 12

Explanation:

[OH^–] = [NaOH] = 0.01 M = 10^-2 M

pOH = –log[OH^–] = –log(10^-2)

pH = 14 – pOH = 12